SECTION 10Mass Transfer
PROBLEM 10.1
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick.
Conditions are fixed so that the gas contains 50% by volume of ammonia at one boundary
of the stagnant layer. The ammonia diffusing to the other boundary is quickly absorbed
and the concentration is negligible at that plane. The temperature is 295 K and the pressure
atmospheric, and under these conditions the diffusivity of ammonia in air is 0.18 cm^2 /s.
Calculate the rate of diffusion of ammonia through the layer.
Solution
See Volume 1, Example 10.1.
PROBLEM 10.2
A simple rectifying column consists of a tube arranged vertically and supplied at the
bottom with a mixture of benzene and toluene as vapour. At the top, a condenser returns
some of the product as a reflux which flows in a thin film down the inner wall of the
tube. The tube is insulated and heat losses can be neglected. At one point in the column,
the vapour contains 70 mol% benzene and the adjacent liquid reflux contains 59 mol%
benzene. The temperature at this point is 365 K. Assuming the diffusional resistance
to vapour transfer to be equivalent to the diffusional resistance of a stagnant vapour
layer 0.2 mm thick, calculate the rate of interchange of benzene and toluene between
vapour and liquid. The molar latent heats of the two materials can be taken as equal. The
vapour pressure of toluene at 365 K is 54.0kN/m^2 and the diffusivity of the vapours is
0 .051 cm^2 /s.
Solution
In this solution, subscripts 1 and 2 refer to the liquid surface and vapour side of the
stagnant layer respectively and subscriptsBandTrefer to benzene and toluene.
If the latent heats are equal and there are no heat losses, there is no net change of phase
across the stagnant layer.
This is an example of equimolecular counter diffusion and:
NADDPA 2 PA 1
/RTL (equation 10.23)whereLDthickness of the stagnant layerD 0 .2mmD 0 .0002 m.
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